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Furnishing a contradiction in my proof involving Lagrange's Theorem

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    furnish.jpg


    3. The attempt at a solution
    I'm trying to show that Case 1 implies that [itex]\tau \in H[/itex], and since [itex]\tau[/itex] was an arbitrarily chosen 3-cycle, then H must contain all 3-cycles, thus contradicting that H has 6 elements.
     
  2. jcsd
  3. Nov 10, 2011 #2

    Deveno

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    first of all, what is σ? it appears to be an arbitrary 3-cycle, but that isn't clear.

    but going on that assumption, note that one of the elements of H is e.

    therefore σ is in Hσ. similarly σσ is in Hσσ.

    but since Hσ = Hσσ, σσ = τkσ, so σ = σσσσ = τkσσσ = τk.

    thus σ is in H, which means <σ> = {σ,σσ,e} is in H.

    but σ is an arbitrary 3-cycle, so....
     
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